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I am trying to prove that given a parametric function of a regular curve in $\mathbb{R}^n$ the curvature of the curve for each $\gamma(t)$ is given by the following expression :

$$k(t)=\frac{\sqrt{||\gamma’(t)||^2||\gamma’’(t)||^2-(\gamma’(t)\cdot\gamma’’(t))^2}}{||\gamma’(t)||^3}$$

I have tried to use an arc-length parametrization of the curve but it leads me nowhere.

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There's no need to use arclength parametrization. We have $\gamma'(t)=s'(t)T(t)$, where $T(t)$ is the unit tangent vector and from the definition of the Frenet frame in $\mathbb{R}^n$ we also have $\gamma''(t)=s''(t)T(t) + s'(t)^2\kappa(t)P_1(t)$, where $P_1(t)$ is the first normal vector in the $n$-dimensional Frenet frame.

Since $\{T,P_1\}$ is an orthonormal set, $\|\gamma'\|^2 = (s')^2$ and $\gamma'\cdot\gamma''=s's''$ and also $\|\gamma''\|^2=(s'')^2+(s')^4\kappa^2$ . The quantity you have under your square root comes out to \begin{equation} (s')^2(s'')^2+(s')^6\kappa^2-(s')^2(s'')^2=(s')^6\kappa^2. \end{equation} The result you are looking for follows.

There are nice formulas for all higher curvatures. You could check out Section 9.1 in Differential Geometry of Curves and Surfaces by Banchoff and Lovett, 2nd edition. (The first edition does not have Chapter 9, so you need 2nd ed. or higher.)

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There is nothing truly insightful: just a lot of computations. Reparameterize $\alpha$ such that $\alpha\circ \Phi$ is arc-length parameterized. This yields:

\begin{equation} \langle \alpha'(\Phi(u))\Phi'(u), \alpha'(\Phi(u))\Phi'(u) \rangle=1 \label{1} \end{equation}

\begin{equation} \langle \alpha''(\Phi(u))\Phi'(u)^2+\alpha'(\Phi(u))\Phi''(u), \alpha'(\Phi(u))\Phi'(u) \rangle=0 \label{2} \end{equation}

The first relation also implies the two following equations:

\begin{equation} \Phi'(u)=\frac{1}{\sqrt{\langle\alpha'(\Phi(u)),\alpha'(\Phi(u))\rangle}}=\frac{1}{|\alpha'(\Phi(u)|} \end{equation}

$$\Phi''(u)=-\frac{1}{2}\frac{1}{\langle \alpha'(\Phi(u)),\alpha'(\Phi(u))\rangle^{\frac{3}{2} }}2\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle\Phi'(u)=-\frac{\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle}{|\alpha'(\Phi(u))|^4}$$

With these four equations, we are ready to prove the main result. First let us introduce the variable $\Gamma=\alpha''(\Phi(u))\Phi'(u)^2+\alpha'(\Phi(u))\Phi''(u)$:

$$ |k|^2=|(\alpha(\Phi(u)))''|^2=|\Gamma|^2=\langle\Gamma,\alpha''(\Phi(u))\rangle\Phi'(u)^2+\langle\Gamma,\alpha'(\Phi(u))\rangle\Phi''(u)$$

By the second equation we have $ \langle\Gamma,\alpha'(\Phi(u))\rangle\Phi''(u)=0$ (because it is a regular parametrization and we may multiply and divide by $\Phi'(u)$). Thus:

$$k^2=\langle\Gamma,\alpha''(\Phi(u))\rangle\Phi'(u)^2=$$ $$\langle\alpha''(\Phi(u)),\alpha''(\Phi(u))\rangle\Phi'(u)^4+\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle\Phi''(u)\Phi'(u)^2=$$ $$\frac{|\alpha''(\Phi(u))|^2}{|\alpha'(\Phi(u))|^4}-\frac{\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle^2}{|\alpha'(\Phi(u))|^6}$$

Taking the square root from both sides yields:

$$k=\frac{\sqrt{|\alpha''(\Phi(u))|^2|\alpha'(\Phi(u))|^2-\langle \alpha'(\Phi(u)),\alpha''(\Phi(u))\rangle^2}}{|\alpha'(\Phi(u))|^3}=$$ $$\frac{\sqrt{|\alpha''(t)|^2|\alpha'(t)|^2-\langle \alpha'(t),\alpha''(t)\rangle^2}}{|\alpha'(t)|^3}$$

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  • $\begingroup$ In addition to not re-using OPs notation, none of the terms and functions are defined α, Φ etc neither their domains, this makes it really hard to read the answer, at least to me. $\endgroup$
    – arkan
    Feb 2, 2023 at 2:14
  • $\begingroup$ @arkan, I will try to help you. $\alpha: I \subset \mathbb{R}\rightarrow \mathbb{R}^n$ is your curve. However, you have no guarantee that $|\alpha'|\equiv 1$, you only know $|\alpha'|$ is not zero anywhere (by definition of a regular curve). Let me define $\Phi^*=\int_0^s |\alpha'|$. This function is invertible and we will call its inverse $\Phi$ (try differentiating $\Phi^*$ and using the Inverse Function Theorem). When you write $\alpha \circ \Phi$ we have $(\alpha\circ \Phi)'\equiv 1$. If any step is not that clear feel free to enquire further. $\endgroup$
    – Kadmos
    Feb 3, 2023 at 9:33
  • $\begingroup$ Yes, thank you so much, this is very helpful. I have questions though. Let $s(t) = \alpha(t)$; My understanding is that the curvature is $s''(\Phi(t))$ it can be shown that $\kappa=s''(\Phi(t))=\frac{\|{T'(t)}\|}{\|{ s'(t)}\|}=\frac{\left\|\frac{s'(t)}{\|s'(t)\|}\right\|}{\|s'(t)\|}=\frac{ \|{s'(t) \times s''(t)}\|}{{\|s'(t)\|}^3}$ . But I don't understand the starting point. When you say "Consider the reparametrization $\alpha\circ \Phi$, this yields to dot product=1; dot product=0" I don't see where those dot product come from. $\endgroup$
    – arkan
    Feb 4, 2023 at 14:58
  • $\begingroup$ After some reverse engineering I guess it comes from: (1) $\|s'(\Phi(t)) \|^2=1$ and (2) $< (s(\Phi(t)))'', s'(\Phi(t))>=0$. I understand (1) as "we must enforce velocity equal 1" in other words $\Phi$ must be an arc length parametrization. As for (2) it is like stating "velocity and acceleration must be perpendicular" which is puzzling, I don't understand why we should enforce such thing to compute the curvature. So yea, I don't understand why the starting point is not the definition of the curvature $k = s''$ $\endgroup$
    – arkan
    Feb 5, 2023 at 2:20
  • $\begingroup$ The problem with your line of reasoning is that $||(\alpha \circ \Phi)’||^2=1$. So you need the chain rule to compute the derivative. Saying that $||\alpha’\circ \Phi||^2=1$ wouldn’t be correct. $\endgroup$
    – Kadmos
    Feb 5, 2023 at 15:48

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